In which case , the product is neither a cyclic ether nor open chain symmetrical ether

C H_{3} - C H = C H - C H_{3} C_{6} H_{5} C O_{3} H - ------ \to

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C H_{3} C H_{2} O N a + C_{2} H_{5} B r \to

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K C N + (C H_{3})_{3} C B r \to

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C_{2} H_{5} O H (E x c e s s) + H_{2} S O_{4} 140^{\circ} C - --- \to

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Solution

The correct option is C $K C N + (C H_{3})_{3} C B r \to$

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